từng câu 1 thôi:v

a) x2-xy+5y-25
 = x(2-y)+ 5(y-2)
 = x(2-y)-5(2-y)
 = (x-5)(2-y)

b: (x-y)(x^2-2x+y)

\(=x^3-2x^2+xy-x^2y+2xy-y^2\)

\(=x^3-2x^2-x^2y+3xy-y^2\)

c: \(\left(x^2-y\right)\left(x+y^2\right)-\left(x-y\right)\left(x^2+xy+y^2\right)\)

\(=x^3+x^2y^2-xy-y^3-\left(x^3-y^3\right)\)

\(=x^2y^2-xy\)

d: \(3x\left(2xy-z\right)-5y\left(x^2-2\right)+3xz\)

\(=6x^2y-3xz-5x^2y+10y+3xz\)

\(=x^2y+10y\)

Bài 2:

a: \(7x^2-14xy+7y^2\)

\(=7\left(x^2-2xy+y^2\right)=7\left(x-y\right)^2\)

b: xy-3x+2y-6

=x(y-3)+2(y-3)

=(x+2)(y-3)

c: \(9x^2+6xy-25+y^2\)

\(=\left(9x^2+6xy+y^2\right)-25\)

\(=\left(3x+y\right)^2-25=\left(3x+y-5\right)\left(3x+y+5\right)\)

Bài 1:

a: 15x+10+4x(3x+2)=0

=>5(3x+2)+4x(3x+2)=0

=>(3x+2)(4x+5)=0

=>\(\left[\begin{array}{l}3x+2=0\\ 4x+5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac23\\ x=-\frac54\end{array}\right.\)

b: \(2x\left(x-6\right)+x^2-36=0\)

=>2x(x-6)+(x-6)(x+6)=0

=>(x-6)(2x+x+6)=0

=>(x-6)(3x+6)=0

=>(x+2)(x-6)=0

=>\(\left[\begin{array}{l}x+2=0\\ x-6=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-2\\ x=6\end{array}\right.\)

a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{4}=\dfrac{y}{-5}=\dfrac{-3x+2y}{-12-10}=\dfrac{55}{-22}=\dfrac{-5}{2}\)

Do đó: \(\left\{{}\begin{matrix}x=\dfrac{-20}{2}=-10\\y=\dfrac{25}{2}\end{matrix}\right.\)

b: Ta có: \(\dfrac{x}{y}=\dfrac{-7}{4}\)

nên \(\dfrac{x}{-7}=\dfrac{y}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{-7}=\dfrac{y}{4}=\dfrac{4x-5y}{-28-20}=\dfrac{72}{-48}=\dfrac{-3}{2}\)

Do đó: \(\left\{{}\begin{matrix}x=\dfrac{21}{2}\\y=\dfrac{-12}{2}=-6\end{matrix}\right.\)

c) \(\dfrac{x}{-3}=\dfrac{y}{8}\)   

⇒\(\dfrac{x^2}{-9}=\dfrac{y^2}{64}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x^2}{-9}=\dfrac{y^2}{64}=-\dfrac{44}{\dfrac{5}{-9+64}}=-\dfrac{44}{\dfrac{5}{55}}=-484\)

a.

\(1-4x^2=\left(1-2x\right)\left(1+2x\right)\)

b.

\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

c.

\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)

\(=\left(x+3\right)^3\)

d.

\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)

e.

\(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y-5\right)\)

f.

\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)

g. 10x(x-y)-6y(y-x)

=10x(x-y)+6y(x-y)

=(x-y)(10x+6y)

h.x²-4x-5

=(x-5)(x+1)

i.x⁴-y⁴ = (x²-y²)(x²+y²)

Mình tính thẳng ra nhé.

a) -A+B-C= -4x^2 + 2xy - 3y^2 + 3y + 7.

b) A+B-(-C)= -5y^2 = 2xy - 4x + 9y + 5.

a: \(\begin{cases}3x-y=5\\ 4x+2y=10\end{cases}\)

=>\(\begin{cases}3x-y=5\\ 2x+y=5\end{cases}\Rightarrow\begin{cases}3x-y+2x+y=5+5\\ 2x+y=5\end{cases}\)

=>\(\begin{cases}5x=10\\ 2x+y=5\end{cases}\Rightarrow\begin{cases}x=2\\ y=5-2\cdot2=5-4=1\end{cases}\)

b: \(\begin{cases}5x+2y=9\\ x+5y=11\end{cases}\Rightarrow\begin{cases}5x+2y=9\\ 5x+25y=55\end{cases}\)

=>\(\begin{cases}5x+25y-5x-2y=55-9\\ x+5y=11\end{cases}\Rightarrow\begin{cases}23y=46\\ x=11-5y\end{cases}\)

=>\(\begin{cases}y=2\\ x=11-5\cdot2=11-10=1\end{cases}\)

c: \(\begin{cases}3x+y=10\\ 4x-3y=9\end{cases}\Rightarrow\begin{cases}9x+3y=30\\ 4x-3y=9\end{cases}\)

=>\(\begin{cases}9x+3y+4x-3y=30+9\\ 3x+y=10\end{cases}\Rightarrow\begin{cases}13x=39\\ y=10-3x\end{cases}\)

=>\(\begin{cases}x=3\\ y=10-3\cdot3=10-9=1\end{cases}\)

d: \(\begin{cases}4x+3y=22\\ 5x+3y=26\end{cases}\Rightarrow\begin{cases}5x+3y-4x-3y=26-22\\ 4x+3y=22\end{cases}\)

=>\(\begin{cases}x=4\\ 3y=22-4x=22-4\cdot4=22-16=6\end{cases}\Rightarrow\begin{cases}x=4\\ y=2\end{cases}\)